Question

How do you integrate ln x / (1+x^2) ?

Answer 1

The integral equals `xln(x/(1 + x^2)) + x- 2arctanx + C`

Explanation:

I'm assuming the question is to integrate `int ln(x/(1 + x^2))dx`. We let `u =ln( x/(1 + x^2))` and `dv = 1`.

`du = (1(1 + x^2) - x(2x))/(1 + x^2)^2 1/(x/(1 +x^2))`

`du = (1 + x^2 - 2x^2)/(1 + x^2)^2 (1 + x^2)/x`

`du = (1 - x^2)/(x(1+ x^2))`

And

`v = x`

We know that integration by parts states that

`int u dv = uv - int vdu`

Therefore

`int ln(x/(1 + x^2))dx = ln(x/(1 +x^2)) x - int x(1 - x^2)/(x(1 +x^2))dx`

`int ln(x/(1 + x^2)) dx = xln(x/(1 + x^2)) - int (1 - x^2)/(1 + x^2) dx`

You're going to want to separate the remaining integral using partial fractions.

`A/1 + B/(1 + x^2) = (1 - x^2)/(1 + x^2)`

`A(1 +x^2) + B = 1 - x^2`

`Ax^2 + A + B = 1 - x^2`

Therefore `A = -1` and `A + B = 1`, thus `B = 2`.

The integral becomes

`int(1- x^2)/(1 + x^2) dx = int -1dx + 2int 1/(1 +x^2) dx`

Recall that `d/dx(arctanx) = 1/(1 + x^2)`.

`int (1 -x^2)/(1 +x^2) dx = -x+ 2arctanx`

Putting the complete integral together we get

`int ln(x/(1 + x^2))dx = xln(x/(1 + x^2)) + x - 2arctanx + C`

Hopefully this helps!